I'm having trouble where to start with this question.. Any help would be helpful.

A projectile is fired with an initial speed of 75.2 m/s at an angle of 34.5° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the velocity of the projectile 1.50 s after firing.

First vertical problem:

Vi = 75.2 sin 34.5 = 42.6 m/s
v = Vi - g t
g = 9.81 m/s^2
so
v = 42.6 - 9.81 t
at the top, v = 0
t = 42.6/9.81 = 4.34 seconds to top

2 t = 8.68 seconds total in air

range = 8.68 * 75.2 cos 34.5

at t = 1.5
v = Vi - g t = 42.6-9.81(1.50)
= 27.9 m/s
u = 75.2 cos 34.5 = 62 m/s

speed = sqrt (u^2+v^2
= sqrt (62^2 + 27.9^2)

tan angle = v/u = 27.9/62
angle = 24.2 degrees above horizontal

Ah, the wonderful world of projectiles! Let's tackle this step by step, shall we?

(a) To find the maximum height reached by the projectile, we have to consider the vertical motion. Now, if you were a projectile, would you aim low and go high, or aim high and go low? Of course, you'd aim high and go low, just like in your social life! So, to find the maximum height, we need to find the vertical component of the initial velocity. That would be 75.2 m/s multiplied by the sine of the launch angle (34.5°). Calculate that, and you'll get the answer!

(b) Now, let's move on to the total time in the air. Well, it's quite simple. The time of flight would be twice the time it takes for the projectile to reach the highest point since it has an equal ascent and descent time. So, we just need to find the time it takes to reach the highest point. But hey, don't worry, the time will fly by just like the projectile! Calculate it using the formula: time = (vertical component of initial velocity) / (acceleration due to gravity).

(c) Ah, the total horizontal distance covered, or as I like to call it, the range! To find this, we need to consider only the horizontal motion. Just like you running away from problems, the horizontal velocity remains constant during the whole flight. Multiply the initial horizontal velocity (75.2 m/s times the cosine of the launch angle (34.5°)) by the total time of flight (which we found in part b). That should give you the range!

(d) Lastly, we need to determine the velocity of the projectile 1.50 seconds after firing. Well, it's a lovely combination of horizontal and vertical motion. The horizontal component of the velocity remains constant throughout the flight, as we mentioned in part c. For the vertical component, you just need to find the acceleration due to gravity, which wants to drag you down! Multiply that by 1.50 seconds and add it to the initial vertical component of the velocity.

And there you have it, my friend! Tackle those problems like a projectile, and they'll be hit out of the park!

To solve this question, we can use the equations of projectile motion. There are four main equations we can use for projectile motion:

1. The horizontal motion equation:
x = v₀x * t
where x is the horizontal distance, v₀x is the initial horizontal velocity, and t is time.

2. The vertical motion equations:
y = v₀y * t - (1/2) * g * t²
where y is the vertical position, v₀y is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s²), and t is time.

3. The equation for horizontal velocity:
v₀x = v₀ * cos(θ)
where v₀x is the initial horizontal velocity, v₀ is the initial speed of the projectile, and θ is the angle of projection.

4. The equation for vertical velocity:
v₀y = v₀ * sin(θ) - g * t
where v₀y is the initial vertical velocity, v₀ is the initial speed of the projectile, θ is the angle of projection, g is the acceleration due to gravity, and t is time.

Now let's solve the parts of the question:

(a) To find the maximum height reached by the projectile, we need to determine when the vertical velocity becomes zero. Using the equation for vertical velocity, we have:
0 = v₀ * sin(θ) - g * t_max
Solve for t_max:
t_max = v₀ * sin(θ) / g
Substitute the given values:
t_max = 75.2 * sin(34.5°) / 9.8
Calculate t_max.

(b) To find the total time in the air, we need to double the time it takes for the projectile to reach its maximum height. Using the equation for t_max found in part (a), we have:
t_total = 2 * t_max
Calculate t_total.

(c) To find the total horizontal distance covered (range), we need to calculate how far the projectile travels horizontally in the time it takes to reach its maximum height. Again, we can use the horizontal motion equation:
x = v₀x * t_max
Calculate x.

(d) To find the velocity of the projectile 1.50 s after firing, we can use the equations for horizontal and vertical velocity. First, calculate the initial horizontal and vertical velocities using the given speed and angle. Then, use the equations for horizontal and vertical velocity to find the velocity 1.50 s after firing.
Calculate v₀x and v₀y.
Use the equations for horizontal and vertical velocity to calculate the velocity 1.50 s after firing.

Remember to double-check your calculations and units to ensure accuracy.